In previous chapters
 agents solve problem by searching
 no knowledge (past experience) is used
 Humans solve problem
 by reusing knowledge and
 new facts concluded under reasoning of knowledge
 even under partially observable environment
 Reasoning is very important
 especially in the world where the results of actions are unknown

The central component of a KB agent
 its memory – knowledge base , or KB
 A KB is a set of representation of facts about the world
 Each of this representation is called a sentence
The sentences are expressed in a language called

 knowledge representation language
(  syntax)

The state diagram, transition operators, search tree

 no longer applicable to build a KB agent because we don ’ t know the results
A new design (design of KB agent) includes:
 A formal language to express its knowledge
 A means of carrying out reasoning in this language
These two elements together constitute
 logic

TELL and ASK
 the way to add new sentences
 and the way to query what sentences the
KB contains
Another main component of KB agent
 the inference engine
 This is to answer the query by ASK
 based on the sentences TELLed in the KB

KB
Tell
Ask KB
Tell
Ask
Inference engine
KB

The query
 may not be answered directly from the sentences
 but usually indirectly
At the initial stage of KB
 Before anything is TELLed
 It contains some background knowledge about the domain
 Example: location(desk, kitchen), …

1.
2.
Each time the agent program is called, it does two things: it TELLs the KB what it perceives it ASKs the KB what action to perform


 reasoning

The process in concluding the action from the sentences in the KB
This reasoning is done logically

 and the action found is proved to be better than all others
(given the agent knows its goals)

P erformance measure
 +1000 for picking up the gold
 -1000 falling into a pit, or being eaten by the wumpus
 -1 for each action taken
 -10 for using up the arrow

E nvironment
 A 4x4 grid of rooms.
 The agent always start in the square [1,1]
 facing to right
 locations of gold and wumpus
 chosen randomly uniformly
 from squares, except [1,1]
 pit
 every square, except [1,1], may be a pit
 with probability 0.2

Actuators (for the agent)
 move forward, turn left, turn right by 90 o
 if there is wall, move forward has no effect
 dies if the agent
 enters a square with pit, or with a live wumpus
 it is safe to enter if the wumpus is dead
 Grab
 pick up an object that is in the same square
 Shoot
 fire an arrow in a straight line in the facing direction
 continues until it hits the wumpus, or a wall
 can be used only 1 time

Sensors – there are 5
 perceive a stench ( 臭味 ) if the agent is
 in the square with wumpus (live or dead)
 and in the directly adjacent squares
 perceive a breeze ( 風 ) if
 in the square directly adjacent to a pit
 perceive a glitter ( 耀眼 )
 in the square containing the gold
 perceive a bump ( 撞 )
 when walks into a wall
 perceive a scream (in all [x,y])
 when the wumpus was killed

The percept is then expressed
 as a state of five elements
 if in a square of stench and breeze only,
 the percept looks like
 [ Stench, Breeze, None, None, None ]
Restriction on the agent
 it cannot perceive the locations adjacent to itself
 but only its own location
 so it is a partially observable environment
 the actions are contingent (nondeterministic)

In most cases,
 the agent can retrieve the gold safely
About 21% of the environments,
 there is no way to succeed
Example,
 the squares around the gold are pits
 or the gold is in a square of pit

Based on the examples,
 thinking logically (rationally) can let the agent take the proven correct actions

The symbols of propositional logic are
 the logical constants True and False

 propositional symbols such as P and Q the logical connectives

,

,

,

,

, and ()
Sentences
 True and False are themselves sentences
 Propositional symbol such P and Q

 is itself a sentence
Wrapping “ ( ) ” around a sentence
 yields a sentence

A sentence





 formed by combining sentences with one of the five logical connectives:

: negation

: conjunction
Antecedent /
Premise

: disjunction

: implication. P

Q : if P then Q

: bidirectional
Conclusion /
Consequent
An atomic sentence
 one containing only one symbol

A literal
 either an atomic sentence
 or a negated one
A complex sentence
 sentences constructed from simpler sentences using logical connectors

The semantics of propositional logic
 simply interpretation of truth values to symbols
 assign True or False to the logical symbols

 e.g., a model m
1
= {P
1,2
= false, P
2,2
= false, P
3,1
= true} the easiest way to summarize the models
 a truth table

Use wumpus world as an example
 where only pits are defined
For each i, j:
 P i,j
= true if there is a pit in [i, j]
 B i,j
= true if there is a breeze in [i, j]
The knowledge base includes (rules)
 There is no pit in [1,1]
R1 :

P
1,1

A square is breezy if and only if
 a pit in a neighboring square

 all squares must be stated now, just show several
Breeze percepts
R2: B
1,1
R3: B
2,1

(P
1,2

(P
1,1

P
2,1

P
2,2
)

P
3,1
)
 from the agent during runtime
Then the KB contains
R4 :

B
1,1
R5 : B
2,1
 the five rules, R
1 to R
5

Based on the existing KB
 a set of facts or rules
 we ASK a query, e.g., is P
2,2 true?
Inference is performed to give the answer
In this case, a truth table is used

For large KB,
 a large amount of memory necessary
 to maintain the truth table
 if there are n symbols in KB
 there are totally 2 n rows (models) in the truth table
 which is O(2 n ), prohibited
 also the time required is not short
 so, we use some rules for inference
 instead of using truth tables

An inference rule
 a rule capturing a certain pattern of inference

 we write α├ β or
List of inference rules
β

All logical equivalences in the following
 can be used as inference rules

We start with R
1
 no pit in [1,2] through R
5 and show how to prove
 From R
2
, apply biconditional elimination

The preceding deviation – called a proof
 a sequence of applications of inference rules
 similar to searching a solution path
 every node is viewed as a sentence
 we are going to find the goal sentence
 hence all algorithms in Ch.3 and Ch.4 can be used
 but it is proved to be NP-complete
 an enormous search tree of sentences can be generated exponentially

This is a complete inference algorithm

 derive all true conclusions from a set of premises
Consider it 
 Agent
 returns from [2,1] to [1,1]
 then goes to [1,2]
 Perception?

 stench no breeze
 All these facts
 added to KB

Similarly to getting R
10
, we know
Apply biconditional elimination to R
3
, then M.P. with R
5
, then
Then apply resolution rule: ¬ P
2,2 with P
2,2 in R
15
We know ¬ P
1,1 is not pit: in R
13 resolves

The previous rules are unit resolution
 where l and m are complementary literals
 m = ¬ l
 the left is called a clause
 a disjunction of literals
 where the left one is called a unit clause
For full resolution rule

Two more examples for resolution
One more technical aspect in resolution


 factoring – removal of multiple copies e.g., resolve (A

B) with (A
 ¬ B) generate (A

A) = A

Resolution rule – a weak point
 applied only to disjunctions of literals
 but most sentences are conjunctive
 so we transform the sentences in CNF
 i.e., expressed as a conjunction of disjunctions of literals
 a sentence in k-CNF has k literals per clause
( l
1,1
 …  l
1,k
)
 … 
( l n,1

…
 l n,k
)

Inference based on resolution

 proof by contradiction, or refutation i.e., show (KB
 ¬ α ) is unsatisfiable
 e.g., To prove α, first assume ¬ α, to see the result can be true or not.
 True  ¬ α = True, False  α = True
The first step, all sentences
 (KB
 ¬ α ) are converted into CNF
 Then apply resolution rules to this CNF
 then see which result it is

In many practical situations,
 resolution is not needed
 reason?
 real-world KB only has Horn clauses
 a disjunction of literals of which


 at most one is positive e.g., ¬ L
1,1
¬ B
1,1

¬ Breeze

B
1,1

P
1,2

P
2,1 is not
 why only 1 positive literal?

Three reasons
1. Horn clause  implication
 premise = conjunction of positive literals
 and conclusion = a single positive literal
 e.g., (L
1,1

Breeze) => B
1,1
 a Horn clause with no positive literals
 (A

B) => false
2. Inference with Horn clauses
 work with Forward and Backward chaining
 both of them
 an easy and natural algorithm

3. Reasoning with Horn clauses
 work in time linear in the size of KB
 cheap for many propositional KB in practice
Forward chaining
 to produce new information
 based on the set of known facts and clauses
 the new information will then be added to KB
 and produce another set of information
 e.g.,
(L
1,1

Breeze) => B
1,1
L
1,1
Breeze
Then? B
1,1 is added

When we want to prove q or find q
Using forward-chaining
 producing new fact continues until
 the query q is added
 no new facts can be generated
 this process runs in linear time
This kind of inference
 called data-driven
 inference starts with the known data
 i.e., clauses (rules) are driven with data

AND-OR graph
 conjunction -multiple links with an arc indication
 disjunction -multiple links without an arc indication

Backward chaining
 the opposite of forward chaining
 A query q is asked
 if it is known already, finished
 otherwise, the algorithm finds all implications that conclude q (the clauses with q as head)
 try to prove all premises in those matched implications
 every premise is then another query q ’
 what is that?
 Prolog matching and unification
 also runs in linear time or fewer than linear time
 only relevant clauses are matched and used

Finding pits and wumpuses, for every [x,y]
 rule for breeze
 rule for stench
B x , y

( P x , y

1

P x , y

1

P x

1 , y

P x

1 , y
)
S x , y

( W x , y

1

W x , y

1

W x

1 , y

W x

1 , y
)
 rules for wumpus
 at least one wumpus: W
1,1

W
1,2
 … 
W
4,4
 and at most one wumpus
 for any two squares, one of them must be wumpus-free

 with n squares, how many sentences? n(n-1)/2 e.g., ¬ W
1,1

¬ W
1,2
 for a 4 x 4 world, there are 105 sentences containing 64 distinct symbols. (each square has S, B, W, P, … )

for every square, the player
 may face 4 different orientation
 Up, Down, Left, Right
 in order to let the player forward
L x , y

FacingRigh t

Forward

L x

1 ,y
 hence the rules are increased to 4 times
 this greatly affect the efficiency
 because too many rules !!

It still works in small domain (4 x 4)
 how about (10 x 10)? and even (100 x 100)?
The main problem
 there are too many distinct propositions to handle
The weakness of Propositional Logic
 lack of expressiveness
 every similar variable must be listed out!!
 we need another powerful device
 First-order logic – discussed in chapter 8